HSC 2015 MX1 Marathon (archive) (1 Viewer)

davidgoes4wce · Jul 12, 2015

Re: HSC 2015 3U Marathon

I have found my own method of doing inverse trig integration. The formula provided in the HSC exams doesn't give us the whole formula to solve the answer.

Not sure if students memorise this formula or is it just me?

The formula above is the one provided in the HSC exams

leehuan · Jul 12, 2015

Re: HSC 2015 3U Marathon

rand_althor said:
$\begin{align*}x&= \frac{1-u}{1+u} \\\frac{\mathrm{d}x}{\mathrm{d}u}&=\frac{\frac{d}{dx}(1-u)\times(1+u)-\frac{d}{dx}(1+u)\times(1-u)}{(1+u)^2} \\&= \frac{-(1+u)-(1-u)}{(1+u)^2} \\&= \frac{-1-u-1+u}{(1+u)^2} \\&= \frac{-2}{(1+u)^2} \\\mathrm{d}x&= \frac{-2}{(1+u)^2}\times\mathrm{d}u \\ \int^1_0\frac{\ln{(1+x)}}{1+x^2}\,\mathrm{d}x&=\int^1_0\frac{\ln{(1+\frac{1-u}{1+u}})}{1+(\frac{1-u}{1+u})^2}\times\frac{-2}{(1+u)^2}\,\mathrm{d}u \\ &= -2\int^1_0 \frac{\ln({\frac{1+u+1-u}{1+u}})}{\frac{(1+u)^2+(1-u)^2}{(1+u)^2}}\times\frac{1}{(1+u)^2}\,\mathrm{d}u \\&= -2\int^1_0 \frac{\ln\frac{2}{(1+u)}}{1+2u+u^2+1-2u+u^2} \,\mathrm{d}u \\ &= -2\int^1_0 \frac{\ln2-\ln(1+u)}{2+2u^2} \,\mathrm{d}u \\&= \int^1_0\frac{\ln(1+u)}{1+u^2} - \frac{\ln2}{1+u^2}\,\mathrm{d}u \\&= \int^1_0\frac{\ln(1+x)}{1+x^2}\,\mathrm{d}x - \ln2\int^1_0\frac{1}{1+u^2}\,\mathrm{d}u \\ &\textup{Not sure what to do with the first integral.} \\ &\textup{Have I made a mistake somewhere?} \end{align*}$

What InteGrand said.

Check whether or not you changed your integration boundaries from the change of variable.

leehuan · Jul 12, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
I have found my own method of doing inverse trig integration. The formula provided in the HSC exams doesn't give us the whole formula to solve the answer.

Not sure if students memorise this formula or is it just me?

The formula above is the one provided in the HSC exams

Some students do memorise the general case. However, for my purposes, sqrt(a^2-x^2) was enough because I could always just rearrange sqrt(A^2-B^2x^2) into that form.

calebwhitaker199 · Jul 12, 2015

Re: HSC 2015 3U Marathon

Could somebody please do question 5 ?

davidgoes4wce · Jul 12, 2015

Re: HSC 2015 3U Marathon

calebwhitaker199 said:
Could somebody please do question 5 ?

do you happen to have the answer? I just want to check if my working out is correct based on the question you asked

Drsoccerball · Jul 12, 2015

Re: HSC 2015 3U Marathon

david your info says you're a uni graduate?

davidgoes4wce · Jul 12, 2015

Re: HSC 2015 3U Marathon

Drsoccerball said:
david your info says you're a uni graduate?

yes I am, is that a bad thing or good thing?

davidgoes4wce · Jul 12, 2015

Re: HSC 2015 3U Marathon

I've always I guess been one of these guys that asks the tutor/teacher/friends questions. In high school I was like this, in uni I was like this. Yes people do get frustrated. And I must admit I did not do 3 Unit maths or 4 Unit maths here before.

VBN2470 · Jul 12, 2015

Re: HSC 2015 3U Marathon

calebwhitaker199 said:
Could somebody please do question 5 ?

You need to draw a diagram to show that $h^2\cot^2{40^{\circ}}+20^2=h^2\cot^2{30^{\circ}}$ , where h is the height of the building. After re-arranging and simplifying you should get

$h=\frac{20}{\sqrt{\cot^2{30^{\circ}}-\cot^2{40^{\circ}}}}$

calebwhitaker199 · Jul 12, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
do you happen to have the answer? I just want to check if my working out is correct based on the question you asked

No i don't have the answer sorry

calebwhitaker199 · Jul 12, 2015

Re: HSC 2015 3U Marathon

VBN2470 said:
You need to draw a diagram to show that $h^2\cot^2{40^{\circ}}+20^2=h^2\cot^2{30^{\circ}}$ , where h is the height of the building. After re-arranging and simplifying you should get

$h=\frac{20}{\sqrt{\cot^2{30^{\circ}}-\cot^2{40^{\circ}}}}$

Thankyou, that evaluates to approximately 27.2m so seems logical

davidgoes4wce · Jul 12, 2015

Re: HSC 2015 3U Marathon

VBN2470 said:
You need to draw a diagram to show that $h^2\cot^2{40^{\circ}}+20^2=h^2\cot^2{30^{\circ}}$ , where h is the height of the building. After re-arranging and simplifying you should get

$h=\frac{20}{\sqrt{\cot^2{30^{\circ}}-\cot^2{40^{\circ}}}}$

Do you mind if I ask and for caleb's sake as well, how do you draw the diagram out?

VBN2470 · Jul 12, 2015

Re: HSC 2015 3U Marathon

Sorry for the bad drawing, but this is how it should look like.

davidgoes4wce · Jul 12, 2015

Re: HSC 2015 3U Marathon

calebwhitaker199 said:
Thankyou, that evaluates to approximately 27.2m so seems logical

Also not sure but I get a h value of 15.91m after entering that into my calculator which seems a bit wrong on my side.

faisalabdul16 · Jul 12, 2015

Re: HSC 2015 3U Marathon

Can confirm VBN2470's working out is correct =)

VBN2470 · Jul 12, 2015

Re: HSC 2015 3U Marathon

I get 15.912.. as well, I think the OP of the question made a calculator error.

calebwhitaker199 · Jul 12, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
Also not sure but I get a h value of 15.91m after entering that into my calculator which seems a bit wrong on my side.

You are right I forgot to square the terms haha

davidgoes4wce · Jul 12, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce · Jul 12, 2015

Re: HSC 2015 3U Marathon

Here is my working out

davidgoes4wce · Jul 12, 2015

Re: HSC 2015 3U Marathon

Just want to check if my working out is correct?

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